20210604, 13:04  #1 
Aug 2020
Guarujá  Brasil
82_{16} Posts 
Divisibility between polynomial expressions
Given the expression
\[x = \frac{n^2 bn}{(2n+k)b}\] For \(b\) an integer coefficient, \(k\) is any integer constant, and index −∞<\(n\)<∞. What is the relationship that must exist between the coefficient \(b\) and constant \(k\), \(x\) will be a sequence of integers generated by the index \(n\)? I suppose it is better to divide the solutions between \(k\)=even and \(k\)=odd. 
20210604, 14:08  #2  
Feb 2017
Nowhere
2^{4}×389 Posts 
Quote:
n/2  (b+k)/4 + (k^2  b^2)/(8*n+4*k4*b) The only way to make the expression a fraction whose denominator remains bounded as n increases without bound, is to take k = b or k = b. If k = b the expression is n/2 + b/2. If k = b the expression is n/2. Either way, the expression will only be an integer for half the integer values of n. 

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